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sasho [114]
4 years ago
5

A 1400-kg car is traveling with a speed of 17.7 m/s. What is the magnitude of the horizontal net force that is required to bring

the car to a halt in a distance of 33.1 m?
Physics
1 answer:
AleksandrR [38]4 years ago
5 0

Answer:

The magnitude of the horizontal net force is 13244 N.

Explanation:

Given that,

Mass of car = 1400 kg

Speed = 17.7 m/s

Distance = 33.1 m

We need to calculate the acceleration

Using equation of motion

v^2-u^2=2as

Where, u = initial velocity

v = final velocity

s = distance

Put the value in the equation

0-(17.7)^2=2a\times 33.1

a=\dfrac{-(17.7)^2}{33.1}

a=-9.46\ m/s^2

Negative sign shows the deceleration.

We need to calculate the net force

Using newton's formula

F = ma

F =1400\times(-9.46)

F=-13244\ N

Negative sign shows the force is opposite the direction of the motion.

The magnitude of the force is

|F| =13244\ N

Hence,  The magnitude of the horizontal net force is 13244 N.

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What is the mass of a stone moving at a speed of 15 meters/second and having a momentum of 7.5 kilogram - meters / seconds
Elanso [62]

Answer:

0.5 kg

Explanation:

The momentum of an object is defined as

p = mv

where

m is the mass

v is the velocity

In this problem we have,

v = 15 m/s is the velocity of the stone

p = 7.5 kg m/s is the momentum

Solving for m, we can find the mass of the stone:

m=\frac{p}{v}=\frac{7.5 kg m/s}{15 m/s}=0.5 kg

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Musya8 [376]

Answer:

4 N

Explanation:

mass = 2 kg

acceleration = 2 m/s^2

Force = mass * acceleration

         = 2 *2

         = 4 N

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3 years ago
A big wheel at a theme park has a diameter of 14m and people on the ride complete one revolution in 24s. calculate the distance
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Explanation:

We'll call the radius r and the diameter d:

We also assume that the riders are at a distance r = d/2 = 7m from the center of the wheel.

The period of the wheel is 24s. The tangent velocity of the wheel (and the riders) will be: (2pi/T)*r = 0.8 m/s (circa).

It means that in 3 minutes (180 seconds) they'll run 0.8 m/s * 180s = 144m.

Hopefully I understood the question. If yes, that's the answer.

7 0
2 years ago
You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l
sdas [7]

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

h= v_it + \frac{1}{2}gt^2\\

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}

<u>vi =  9.81 m/s</u>

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

h= v_it + \frac{1}{2}gt^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

<u>t = 1.13 s</u>

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

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